The sum of the focal distances of the point $\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$ on the ellipse $9x^2+4y^2=36$ is

An ellipse with its minor and major axes parallel to the coordinate axes passes through $(0,0)$,$(1,0)$,and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is

If a tangent to the ellipse $x^{2}+4y^{2}=4$ meets the tangents at the extremities of its major axis at $B$ and $C$,then the circle with $BC$ as diameter passes through the point:

Let each of the two ellipses $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1, (A < B)$ have eccentricity $\frac{4}{5}$. Let the lengths of the latus recta of $E_1$ and $E_2$ be $\ell_1$ and $\ell_2$, respectively, such that $2\ell_1^2 = 9\ell_2$. If the distance between the foci of $E_1$ is $8$, then the distance between the foci of $E_2$ is:

Two sets $A$ and $B$ are defined as follows:
$A = \{ (a,b) \in R \times R : |a - 5| < 1 \text{ and } |b - 5| < 1 \}$
$B = \{ (a,b) \in R \times R : 4(a - 6)^2 + 9(b - 5)^2 \le 36 \}$
Then:

Let $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ be an ellipse. Ellipses $E_i$ are constructed such that their centres and eccentricities are the same as that of $E_1$,and the length of the minor axis of $E_i$ is the length of the major axis of $E_{i+1}$ $(i \geq 1)$. If $A_i$ is the area of the ellipse $E_i$,then $\frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right)$ is equal to . . . . . . .